Given an array arr[] of size N, and an integer K representing a digit, the task is to print the given array in increasing order according to the increasing frequency of the digit K in the array elements.
Examples:
Input: arr[] = {15, 66, 26, 91}, K = 6
Output: 15 91 26 66
Explanation:
The frequency of digit 6 in array elements is {0, 2, 1, 0}. The elements in increasing order of frequency of the digit 6 in {15, 91, 26, 66}.Input: arr[] = {20, 21, 0}, K = 0
Output: 21 20 0
Explanation:
The frequency of digit 0 in array elements is {1, 0, 1}. The elements in increasing order of frequency of the digit 0 is {21, 20, 0}.
Approach: The idea is to count the occurrences of the digit K for each element of the array and sort the array according to it. Follow the steps below to solve the problem:
- Initialize a multimap, say mp, to store occurrences of K in each element in sorted order.
- Traverse the given array arr[] using the variable i and perform the following:
- Store the occurrences of digit K in arr[i] in a variable cnt.
- Insert a pair of cnt and arr[i] in mp.
- Print the array elements in mp to get the required sorted order.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count the occurrences// of digit K in an elementint countOccurrences(int num, int K){ // If num and K are both 0 if (K == 0 && num == 0) return 1; // Initialize count as 0 int count = 0; // Count for occurrences of digit K while (num > 0) { if (num % 10 == K) count++; num /= 10; } // Return the count return count;}// Function to print the given array// in increasing order of the digit// K in the array elementsvoid sortOccurrences(int arr[], int N, int K){ // Stores the occurrences of K // in each element multimap<int, int> mp; // Traverse the array for (int i = 0; i < N; i++) { // Count the frequency // of K in arr[i] int count = countOccurrences( arr[i], K); // Insert elements in mp mp.insert(pair<int, int>( count, arr[i])); } // Print the elements in the map, mp for (auto& itr : mp) { cout << itr.second << " "; }}// Driver Codeint main(){ int arr[] = { 15, 66, 26, 91 }; int K = 6; int N = sizeof(arr) / sizeof(arr[0]); // Function Call sortOccurrences(arr, N, K); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to count the occurrences // of digit K in an element static int countOccurrences(int num, int K) { // If num and K are both 0 if (K == 0 && num == 0) return 1; // Initialize count as 0 int count = 0; // Count for occurrences of digit K while (num > 0) { if (num % 10 == K) count++; num /= 10; } // Return the count return count; } // Pair class static class Pair { int idx; int freq; Pair(int idx, int freq) { this.idx = idx; this.freq = freq; } } // Function to print the given array // in increasing order of the digit // K in the array elements static void sortOccurrences(int arr[], int N, int K) { // Stores the Pair with index // and occurrences of K of each element Pair mp[] = new Pair[N]; // Traverse the array for (int i = 0; i < N; i++) { // Count the frequency // of K in arr[i] int count = countOccurrences(arr[i], K); // Insert Pair in mp mp[i] = new Pair(i, count); } // sort the mp in increasing order of freq // if freq equal then according to index Arrays.sort(mp, (p1, p2) -> { if (p1.freq == p2.freq) return p1.idx - p2.idx; return p1.freq - p2.freq; }); // Print the elements in the map, mp for (Pair p : mp) { System.out.print(arr[p.idx] + " "); } } // Driver Code public static void main(String[] args) { int arr[] = { 15, 66, 26, 91 }; int K = 6; int N = arr.length; // Function Call sortOccurrences(arr, N, K); }}// This code is contributed by Kingash. |
Python3
# Python program for the above approach# Function to count the occurrences# of digit K in an elementdef countOccurrences( num, K): # If num and K are both 0 if (K == 0 and num == 0): return 1 # Initialize count as 0 count = 0 # Count for occurrences of digit K while (num > 0): if (num % 10 == K): count += 1 num //= 10 # Return the count return count# Function to print the given array# in increasing order of the digit# K in the array elementsdef sortOccurrences(arr, N, K): # Stores the occurrences of K # in each element mp = [] # Traverse the array for i in range(N): # Count the frequency # of K in arr[i] count = countOccurrences(arr[i], K) # Insert elements in mp mp.append([count, arr[i]]) # Print the elements in the map, mp mp = sorted(mp) for itr in mp: print(itr[1], end = ' ')# Driver Codearr = [ 15, 66, 26, 91 ]K = 6N = len(arr)# Function CallsortOccurrences(arr, N, K);# This code is contributed by rohitsingh07052. |
C#
// C# program for the above approachusing System;public class GFG { // Function to count the occurrences // of digit K in an element static int countOccurrences(int num, int K) { // If num and K are both 0 if (K == 0 && num == 0) return 1; // Initialize count as 0 int count = 0; // Count for occurrences of digit K while (num > 0) { if (num % 10 == K) count++; num /= 10; } // Return the count return count; } // Pair class class Pair : IComparable<Pair> { public int idx; public int freq; public Pair(int idx, int freq) { this.idx = idx; this.freq = freq; } public int CompareTo(Pair p) { if (this.freq == p.freq) return this.idx - p.idx; return this.freq - p.freq; } } // Function to print the given array // in increasing order of the digit // K in the array elements static void sortOccurrences(int []arr, int N, int K) { // Stores the Pair with index // and occurrences of K of each element Pair []mp = new Pair[N]; // Traverse the array for (int i = 0; i < N; i++) { // Count the frequency // of K in arr[i] int count = countOccurrences(arr[i], K); // Insert Pair in mp mp[i] = new Pair(i, count); } // sort the mp in increasing order of freq // if freq equal then according to index Array.Sort(mp); // Print the elements in the map, mp foreach (Pair p in mp) { Console.Write(arr[p.idx] + " "); } } // Driver Code public static void Main(String[] args) { int []arr = { 15, 66, 26, 91 }; int K = 6; int N = arr.Length; // Function Call sortOccurrences(arr, N, K); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program for the above approach// Function to count the occurrences// of digit K in an elementfunction countOccurrences( num, K){ // If num and K are both 0 if (K == 0 && num == 0) return 1 // Initialize count as 0 count = 0 // Count for occurrences of digit K while (num > 0){ if (num % 10 == K) count += 1 num = Math.floor(num/10) } // Return the count return count}// Function to print the given array// in increasing order of the digit// K in the array elementsfunction sortOccurrences(arr, N, K){ // Stores the occurrences of K // in each element mp = [] // Traverse the array for(let i=0;i<N;i++){ // Count the frequency // of K in arr[i] let count = countOccurrences(arr[i], K) // Insert elements in mp mp.push([count, arr[i]]) } // Print the elements in the map, mp mp.sort() for([itr1,itr2] of mp) document.write(itr2,' ')}// Driver Codelet arr = [ 15, 66, 26, 91 ]let K = 6let N = arr.length// Function CallsortOccurrences(arr, N, K)// This code is contributed by shinjanpatra</script> |
Output:
15 91 26 66
Time Complexity: O(N*log10N)
Auxiliary Space: O(N)
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